What is the solution for (x)^2y^{''}+3(x)y^'-3y=(x)^3 ?

The solution for (x)^2y^{''}+3(x)y^'-3y=(x)^3 is y=c_{1}x+(c_{2})/(x^3)+(x^3)/(12)

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(x)^2y^{''}+3(x)y^'-3y=(x)^3

(x)^{2} y^{^{''}} + 3 (x) y^{^{'}} - 3 y = (x)^{3}

y = c_{1} x + \frac{c _{2}}{x ^{3}} + \frac{x ^{3}}{12}

Solution steps

x^{2} y^{''} (x) + 3 (x) y^{'} (x) - 3 y = x^{3}

Solve linear ODE:

y = c_{1} x + \frac{c _{2}}{x ^{3}} + \frac{x ^{3}}{12}

y = c_{1} x + \frac{c _{2}}{x ^{3}} + \frac{x ^{3}}{12}

d er i v a t i v e \frac{1}{t ^{2}}

\int_{0}^{1} x \cdot e^{x} d x

d er i v a t i v e y = x - x^{3}

\int \frac{x ^{2} - 5}{( x + 1 ) ^{3}} d x

\int_{0}^{π} 20 sin^{4} (x) d x

What is the solution for (x)^2y^{''}+3(x)y^'-3y=(x)^3 ?
The solution for (x)^2y^{''}+3(x)y^'-3y=(x)^3 is y=c_{1}x+(c_{2})/(x^3)+(x^3)/(12)